na +i2
To balance the chemical equation Na + I2 -> NaI, you need to ensure that there is an equal number of atoms of each element on both sides of the equation. In this case, you would need to put a coefficient of 2 in front of NaI to balance the equation.
Sodium + Iodine ----> Sodium Iodide2 Na + I2 ----> 2 NaI
The chemical formula of sodium iodide is NaI.
2KI+MnO_2+3H_2 SO_4→2I+2KHSO_4+MnSO_4+2H_2 O
By this, I assume you mean, the chemical reaction. If this is the case, the answer is 2I +2NaCl
To balance the chemical equation Na + I2 -> NaI, you need to ensure that there is an equal number of atoms of each element on both sides of the equation. In this case, you would need to put a coefficient of 2 in front of NaI to balance the equation.
When sodium metal is combined with iodine gas, an oxidation-reduction reaction occurs. Sodium loses and electron to form the sodium cation, and iodide gains an electron to form iodide. The resulting compound is NaI.
Sodium + Iodine ----> Sodium Iodide2 Na + I2 ----> 2 NaI
Na and Zn are correct because they are elements. Na is monovalent so NaI is correct. Zn is divalent and so ZnI is incorrect and should be ZnI2 . So if we have ZnI2 we need 2 Na to remove the I2 . So balanced it becomes: 2Na + ZnI2 ----> 2NaI + Zn
The bond formed between the two is an Ionic bond. You can tell by using the difference in the elements electronegative, or just know the general trend that a metal and nonmetal form an ionic bond.
Sodium + Bromine ----> Sodium bromide2 Na + Br2 ----> 2 NaBr
The balanced chemical equation for the reaction between Br₂ and NaI is: Br₂ + 2NaI -> 2NaBr + I₂ In this reaction, two moles of sodium iodide (NaI) react with one mole of bromine (Br₂) to form two moles of sodium bromide (NaBr) and one mole of iodine (I₂).
The balanced chemical equation for this reaction is: 2NaI + MnO2 + 2H2SO4 -> 2NaHSO4 + MnSO4 + 2H2O + I2 From the balanced equation, 1 mol of NaI gives 1 mol of I2. First, find the number of moles of NaI and MnO2, then determine the limiting reactant. Finally, calculate the moles of I2 formed based on the limiting reactant and convert to grams using the molar mass of I2.
When sodium hypochlorite (NaClO) reacts with potassium iodide (KI), it forms potassium chloride (KCl), sodium iodide (NaI), and elemental iodine (I2). This reaction can be represented by the equation: 3NaClO + 2KI → KCl + NaCl + NaI + I2.
Probably an N Dubz song
Solid sodium iodide reacts with water to form sodium ions and iodide ions. No further reactions occur because this compound is a salt of a strong acid and a strong base. Thus, the overall reaction is NaI(s)=> Na+ + I-
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